The feasible region has corner points $(0,0)$, $(4,0)$ and $(0,4)$. Evaluate $Z=3x+4y$: at $(0,0)$, $Z=0$; at $(4,0)$, $Z=12$; at $(0,4)$, $Z=16$. Hence the maximum value is $16$ at $(0,4)$.
Maximum $Z=16$ at $(0,4)$.
The corner points are $(0,0)$, $(4,0)$, $(2,3)$ and $(0,4)$. The values of $Z=-3x+4y$ are $0$, $-12$, $6$ and $16$, respectively. Therefore the minimum is $-12$ at $(4,0)$.
Minimum $Z=-12$ at $(4,0)$.
The corner points are $(0,0)$, $(2,0)$, $(0,3)$ and the intersection of $3x+5y=15$ and $5x+2y=10$. Solving the two equations gives $x=20/19$, $y=45/19$. The objective values are $0$, $10$, $9$ and $5(20/19)+3(45/19)=235/19$. Hence the maximum is $235/19$ at $\left(20/19,45/19\right)$.
Maximum $Z=\dfrac{235}{19}$ at $\left(\dfrac{20}{19},\dfrac{45}{19}\right)$.
The relevant corner points of the feasible region are $(0,2)$, $\left(\dfrac32,\dfrac12\right)$ and $(3,0)$. The values of $Z=3x+5y$ are $10$, $3(3/2)+5(1/2)=7$ and $9$. Therefore the minimum is $7$ at $\left(\dfrac32,\dfrac12\right)$.
Minimum $Z=7$ at $\left(\dfrac32,\dfrac12\right)$.
The corner points are $(0,0)$, $(5,0)$, $(4,3)$ and $(0,5)$. Evaluating $Z=3x+2y$ gives $0$, $15$, $18$ and $10$. Hence the maximum value is $18$ at $(4,3)$.
Maximum $Z=18$ at $(4,3)$.
Since one constraint is $x+2y\ge6$, the objective $Z=x+2y$ cannot be less than $6$. On the line segment joining $(0,3)$ and $(6,0)$, we have $x+2y=6$, $x,y\ge0$, and $2x+y\ge3$. Therefore every point of this segment is feasible and gives $Z=6$, so the minimum occurs at more than two points.
Minimum $Z=6$ at every point of the segment $x+2y=6$, $0\le x\le6$, $y=3-\dfrac{x}{2}$.
The feasible corner points are $(60,0)$, $(120,0)$, $(60,30)$ and $(40,20)$. The values of $Z=5x+10y=5(x+2y)$ are $300$, $600$, $600$ and $400$. Hence the minimum is $300$ at $(60,0)$. The maximum is $600$ and is attained along the feasible edge $x+2y=120$.
Minimum $Z=300$ at $(60,0)$; maximum $Z=600$ at every feasible point on $x+2y=120$ from $(120,0)$ to $(60,30)$.
The feasible corner points are $(0,50)$, $(20,40)$, $(50,100)$ and $(0,200)$. The values of $Z=x+2y$ are $100$, $100$, $250$ and $400$. Therefore the minimum is $100$ along the feasible edge from $(0,50)$ to $(20,40)$, and the maximum is $400$ at $(0,200)$.
Minimum $Z=100$ at every feasible point on $x+2y=100$ from $(0,50)$ to $(20,40)$; maximum $Z=400$ at $(0,200)$.
The constraints allow points such as $(3,y)$ for all sufficiently large $y$. For these points, $Z=-3+2y$, which increases without bound as $y$ increases. Hence the objective function has no maximum value.
No maximum value exists; the feasible region is unbounded in a direction in which $Z$ increases.
The first inequality gives $x-y\le-1$, i.e. $y\ge x+1$. The second gives $-x+y\le0$, i.e. $y\le x$. These two conditions cannot hold simultaneously. Hence the feasible region is empty and the problem has no maximum solution.
There is no feasible solution, so no maximum exists.